SOLUTION: 3x+4y=5, 2x+y=1...If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. This is what I ha

Algebra ->  Linear-equations -> SOLUTION: 3x+4y=5, 2x+y=1...If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. This is what I ha      Log On


   

Question 592587: 3x+4y=5, 2x+y=1...If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this.
This is what I have......
3x+4y=5, 2x+y=1
3x+4y=5
3x-3x+4y=5-3x
4y=5-3x
4y=3x+5
y=3x+5/4
2x+y=1
2x-2x+y=1-2x
y=1-2x
1y=-2x+1/1
y=-2x+1 no solution
What am I doing wrong here?
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Your first error is at the step where you went from:

Correct

to

Incorrect (Sign error)

Should be:



Then you made another error when you divided by 4. You only divided one of the RHS terms by 4, so instead of:



It should have been (counting correcting the sign error from before)



I'll let you take it from there. Write back if you are still having trouble.

John

My calculator said it, I believe it, that settles it
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